Let $h$ be a vector-valued function defined by $h(t)=\left(-\dfrac3{t+2},e^{3t}\right)$. Find $h$ 's second derivative $h''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-\dfrac{3}{(t+2)^3},9e^{3t}\right)$ (Choice B) B $\left(-\dfrac{6}{(t+2)^3},9e^{3t}\right)$ (Choice C) C $\left(\dfrac{6}{(t+2)^2},6e^{2t}\right)$ (Choice D) D $\left(\dfrac{3}{(t+2)^2},3e^{3t}\right)$
Solution: We are asked to find the second derivative of $h$. This means we need to differentiate $h$ twice. In other words, we differentiate $h$ once to find $h'$, and then differentiate $h'$ (which is a vector-valued function as well) to find $h''$. Recall that $h(t)=\left(-\dfrac3{t+2},e^{3t}\right)$. Therefore, $h'(t)=\left(\dfrac{3}{(t+2)^2},3e^{3t}\right)$. Now let's differentiate $h'(t)=\left(\dfrac{3}{(t+2)^2},3e^{3t}\right)$ to find $h''$. $h''(t)=\left(-\dfrac{6}{(t+2)^3},9e^{3t}\right)$ In conclusion, $h''(t)=\left(-\dfrac{6}{(t+2)^3},9e^{3t}\right)$.